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प्रश्न
Find the sum of all 3 - digit natural numbers which are divisible by 13.
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उत्तर
In the given problem, we need to find the sum of terms for different arithmetic progressions. So, here we use the following formula for the sum of n terms of an A.P.,
`S_n = n/2 [2a +(n -1)d]`
Where; a = first term for the given A.P.
d = common difference of the given A.P.
n = number of terms
All 3 digit natural number which is divisible by 13
So, we know that the first 3 digit multiple of 13 is 104 and the last 3 digit multiple of 13 is 988.
Also, all these terms will form an A.P. with the common difference of 13.
So here,
First term (a) = 104
Last term (l) = 988
Common difference (d) = 13
So, here the first step is to find the total number of terms. Let us take the number of terms as n.
`Now, as we know,
`a_n = a + (n - 1)d`
So, for the last term,
988 = 104 + (n - 1)13
988 = 104 + 13n - 13
988 = 91 + 13n
Further simplifying,
`n = (988 - 91)/13`
`n = 897/13`
n = 69
Now, using the formula for the sum of n terms, we get
`S_n = 69/2 [2(104) + (69 - 1)3]`
`= 69/2 [208 + (68)13]`
`= 69/2 (208 + 884)`
On further simplification, we get,
`S_n = 69/2 (1092)`
= 69 (546)
= 37674
Therefore, the sum of all the 3 digit multiples of 13 is `S_n = 37674`
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